Aus Online Mathematik Brückenkurs 2

Version vom 12:58, 14. Okt. 2008 (bearbeiten) Tek (Diskussion | Beiträge) K ← Zum vorherigen Versionsunterschied		Version vom 12:51, 10. Mär. 2009 (bearbeiten) (rückgängig) Tekbot (Diskussion | Beiträge) K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
	The ball hits the ground when its height is zero, i.e. when		The ball hits the ground when its height is zero, i.e. when
-	{{Displayed math||<math>h(t) = 10-\frac{9\textrm{.}82}{2}t^{2} = 0\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>h(t) = 10-\frac{9\textrm{.}82}{2}t^{2} = 0\,\textrm{.}</math>}}
	This quadratic equation has the solutions		This quadratic equation has the solutions
-	{{Displayed math||<math>t=\pm\sqrt{\frac{2\cdot 10}{9\textrm{.}82}}\,,</math>}}	+	{{Abgesetzte Formel||<math>t=\pm\sqrt{\frac{2\cdot 10}{9\textrm{.}82}}\,,</math>}}
	where the positive root is the time when the ball hits the ground.		where the positive root is the time when the ball hits the ground.
Zeile 11:		Zeile 11:
	We obtain the ball's speed as a function of time as the time derivative of the height,		We obtain the ball's speed as a function of time as the time derivative of the height,
-	{{Displayed math||<math>v(t) = h'(t) = \frac{d}{dt}\,\Bigl(10-\frac{9\textrm{.}82}{2}t^2\Bigr) = -9\textrm{.}82t\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>v(t) = h'(t) = \frac{d}{dt}\,\Bigl(10-\frac{9\textrm{.}82}{2}t^2\Bigr) = -9\textrm{.}82t\,\textrm{.}</math>}}
	If we substitute the time when the ball hits the ground, we obtain the ball's speed at that instant,		If we substitute the time when the ball hits the ground, we obtain the ball's speed at that instant,
-	{{Displayed math||<math>\begin{align}	+	{{Abgesetzte Formel||<math>\begin{align}
	v\Bigl(\sqrt{\frac{2\cdot 10}{9\textrm{.}82}}\,\Bigr)		v\Bigl(\sqrt{\frac{2\cdot 10}{9\textrm{.}82}}\,\Bigr)
	&= -9\textrm{.}82\cdot\sqrt{\frac{2\cdot 10}{9\textrm{.}82}}\\[5pt]		&= -9\textrm{.}82\cdot\sqrt{\frac{2\cdot 10}{9\textrm{.}82}}\\[5pt]

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Version vom 12:51, 10. Mär. 2009

The ball hits the ground when its height is zero, i.e. when

\displaystyle h(t) = 10-\frac{9\textrm{.}82}{2}t^{2} = 0\,\textrm{.}

This quadratic equation has the solutions

\displaystyle t=\pm\sqrt{\frac{2\cdot 10}{9\textrm{.}82}}\,,

where the positive root is the time when the ball hits the ground.

We obtain the ball's speed as a function of time as the time derivative of the height,

\displaystyle v(t) = h'(t) = \frac{d}{dt}\,\Bigl(10-\frac{9\textrm{.}82}{2}t^2\Bigr) = -9\textrm{.}82t\,\textrm{.}

If we substitute the time when the ball hits the ground, we obtain the ball's speed at that instant,

\displaystyle \begin{align} v\Bigl(\sqrt{\frac{2\cdot 10}{9\textrm{.}82}}\,\Bigr) &= -9\textrm{.}82\cdot\sqrt{\frac{2\cdot 10}{9\textrm{.}82}}\\[5pt] &= -\sqrt{9\textrm{.}82^2\cdot\frac{2\cdot 10}{9\textrm{.}82}}\\[5pt] &= \sqrt{9\textrm{.}82\cdot 2\cdot 10}\\[5pt] &= -\sqrt{196\textrm{.}4}\\[5pt] &\approx -14\textrm{.}0\,\textrm{.} \end{align}

The minus sign indicates that the speed is directed downwards, and the ball's speed is therefore 14.0 m/s.